If the sum of the slopes of the lines given by $x^{2} - 4 p x y + 8 y^{2} = 0$ is three times their product then $p$ has the value

\frac{1}{4}

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4

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3

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\frac{3}{4}

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Solution

The correct option is D $\frac{3}{4}$
Given, $m_{1} + m_{2} = 3 m_{1} m_{2}$
$\Rightarrow \frac{- 2 h}{b} = \frac{3 a}{b}$
$\Rightarrow - 2 h = 3 a$
$\Rightarrow - (- 4 p) = 3$
$\Rightarrow p = \frac{3}{4}$

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